## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 11 - Chemical Kinetics: Rates of Reactions - Problem Solving Practice 11.1 - Page 479: a

#### Answer

(i)$$Rate = 3.34 \times 10^{-7} mol L^{-1} s^{-1}$$ (ii)$$Rate = 3.80 \times 10^{-7} mol L^{-1} s^{-1}$$ (iii)$$Rate = 4.768 \times 10^{-7} mol L^{-1} s^{-1}$$

#### Work Step by Step

1. Calculate $\Delta[Cv^+]$ and $\Delta t$: (i) $$\Delta[Cv^+] = 0.793 \times 10^{-5} - 1.460 \times 10^{-5} = -6.67 \times 10^{-6}$$ $$\Delta t = 60.0 s - 40.0 s = 20.0 s$$ (ii) $$\Delta[Cv^+] = 0.429 \times 10^{-5} - 2.710 \times 10^{-5} = -2.281 \times 10^{-5}$$ $$\Delta t = 80.0 s - 20.0 s = 60.0 s$$ (iii) $$\Delta[Cv^+] = 0.232 \times 10^{-5} - 5.000 \times 10^{-5} = -4.768 \times 10^{-5}$$ $$\Delta t = 100.0 s - 0.0 s = 100.0 s$$ 2. Calculate each rate: (i)$$Rate = -\frac{-6.67 \times 10^{-6}}{20.0} = 3.34 \times 10^{-7} mol L^{-1} s^{-1}$$ (ii)$$Rate = -\frac{-2.281 \times 10^{-5}}{60.0} = 3.80 \times 10^{-7} mol L^{-1} s^{-1}$$ (iii)$$Rate = -\frac{-4.768 \times 10^{-5}}{100.0} = 4.768 \times 10^{-7} mol L^{-1} s^{-1}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.